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an investigation of dual stator winding induction machines

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There are two unknown variables in (2.8). One equation c<strong>an</strong>not solve two unknowns,<br />

hence the <strong>stator</strong> yoke flux is chosen as the second objective function in this design. The<br />

<strong>stator</strong> yoke flux corresponds to the integral <strong>of</strong> the air gap flux density, for the <strong>dual</strong>-<br />

<strong>winding</strong> design:<br />

∫<br />

θ<br />

[ ( θ ) − K cos(<br />

3θ<br />

+ δ ) ]<br />

φ y = rlB cos dθ<br />

(2.9)<br />

2 / 6<br />

2<br />

0<br />

1<br />

1<br />

where, r is the me<strong>an</strong> air gap radius <strong>an</strong>d l is the <strong>stator</strong> core length. After the integration,<br />

the <strong>stator</strong> yoke flux c<strong>an</strong> be written as:<br />

⎡ K1<br />

⎤<br />

φ y 2 / 6 = rlB ⎢sin(<br />

θ ) − sin(<br />

3θ<br />

+ δ1<br />

) ⎥⎦<br />

(2.10)<br />

⎣ 3<br />

2<br />

At the peak point <strong>of</strong> yoke flux, the derivative <strong>of</strong> φ y2<br />

6 with respect to <strong>an</strong>gle θ is<br />

equal to zero, which c<strong>an</strong> be expressed as:<br />

φ<br />

d y2<br />

/ 6<br />

dθ<br />

[ ( θ ) − K cos(<br />

3θ<br />

+ ) ] = 0<br />

cos 2<br />

1<br />

1<br />

= rlB<br />

δ<br />

π<br />

When θ = <strong>an</strong>d 1 0<br />

2<br />

= δ , the peak value <strong>of</strong> the <strong>stator</strong> yoke flux c<strong>an</strong> be found as:<br />

51<br />

(2.11)<br />

⎛ K1<br />

⎞<br />

φ y2<br />

/ 6 = rlB2⎜1<br />

+ ⎟<br />

(2.12)<br />

⎝ 3 ⎠<br />

The peak value <strong>of</strong> the flux in the yoke for the 4-pole machine c<strong>an</strong> be written as:<br />

rlB 4 φ y 4 =<br />

(2.13)<br />

2<br />

To maintain the same level <strong>of</strong> saturation, the peak yoke flux value <strong>of</strong> the 2/6 pole<br />

design <strong>an</strong>d the 4-pole design should be the same, which c<strong>an</strong> be expressed as:<br />

⎛ K1<br />

⎞ rlB 4<br />

rlB 2⎜1<br />

+ ⎟ =<br />

(2.14)<br />

⎝ 3 ⎠ 2

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